I know that the following two statements are correct.
- Every open set of $\mathbb{R}$ be written as countable union of disjoint open intervals ( including open rays and $\mathbb{R}$).
- Every open set of $\mathbb{R}$ be written as countable union of open bounded intervals.
Then is it true that every open set of $\mathbb{R}$ be written as countable union of disjoint open bounded intervals? I think following a similar argument as statement 1, we only need to show open rays and $\mathbb{R}$ satisfy this property.
Best Answer
Consider just $\mathbb R$ itself. Suppose that it was a union of bounded open intervals $\mathbb{R} = \bigcup_{i \in I} ( a_i , b_i )$ where $a_i < b_i$ for each $i \in I$ (I'm not so concerned with the countability of $I$). Picking any $i \in I$, consider $a_{i}$. This is not in $( a_{i} , b_{i} )$, so there must be a $j \neq i$ such that $a_{i} \in ( a_j , b_j )$. But then as $a_j < a_i < b_j$ it is easy to check that $( a_{i} , b_i ) \cap ( a_j , b_j ) \neq \varnothing$. (So $\mathbb{R}$ is not the union of any family of pairwise disjoint bounded open intervals.)